# Solve the differential equations (t - 1)3 ds/dt + 4(t

Solve the differential equations $$\displaystyle{\left({t}-{1}\right)}{3}{d}\frac{{s}}{{\left.{d}{t}\right.}}+{4}{\left({t}-{1}\right)}{2}{s}={t}+{1},{t}{>}{1}$$

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censoratojk
Step 1
Given: $$\displaystyle{\left({t}-{1}\right)}^{{{3}}}{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{4}{\left({t}-{1}\right)}^{{{2}}}\ {s}={t}+{1};{t}{>}{1}$$
Step 2
Calculation:
$$\displaystyle{\left({t}-{1}\right)}^{{{3}}}{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{4}{\left({t}-{1}\right)}^{{{2}}}{s}={t}+{1};{t}{>}{1}$$
$$\displaystyle{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{\frac{{{4}{\left({t}-{1}\right)}^{{{2}}}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}{s}={\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}$$
$$\displaystyle{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{\frac{{{4}}}{{{\left({t}-{1}\right)}}}}\ {s}={\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}$$
Compare the differential equation with $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{P}{y}={Q}$$
Here $$\displaystyle{P}={\frac{{{4}}}{{{\left({t}-{1}\right)}^{{{2}}}}}}\ \text{and}{Q}={\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}$$
Solution of given differential equation is $$\displaystyle{y}.{\left({I}.{F}\right)}=\int{Q}.{\left({I}.{F}\right)}{\left.{d}{x}\right.}+{c}$$
Now, $$\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}=\int{\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}.{\left({t}-{1}\right)}^{{{4}}}{\left.{d}{t}\right.}+{c}$$
$$\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}=\int{\left({t}+{1}\right)}.{\left({t}-{1}\right)}{\left.{d}{t}\right.}+{c}$$
$$\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}=\int{\left({t}^{{{2}}}-{1}\right)}{\left.{d}{t}\right.}+{c}$$
$$\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}={\frac{{{t}^{{{3}}}}}{{{3}}}}-{t}+{c}$$
###### Not exactly what youâ€™re looking for?
godsrvnt0706
$$\displaystyle{\left({t}-{1}\right)}^{{{3}}}{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{4}{\left({t}-{1}\right)}^{{{2}}}\ {s}={t}+{1};{t}{>}{1}$$
$$\displaystyle{\left({t}-{1}\right)}^{{{3}}}{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{4}{\left({t}-{1}\right)}^{{{2}}}{s}={t}+{1};{t}{>}{1}$$
$$\displaystyle{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{\frac{{{4}{\left({t}-{1}\right)}^{{{2}}}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}{s}={\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}$$
$$\displaystyle{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{\frac{{{4}}}{{{\left({t}-{1}\right)}}}}\ {s}={\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{P}{y}={Q}$$
$$\displaystyle{P}={\frac{{{4}}}{{{\left({t}-{1}\right)}^{{{2}}}}}}\ \text{and}{Q}={\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}$$
$$\displaystyle{y}.{\left({I}.{F}\right)}=\int{Q}.{\left({I}.{F}\right)}{\left.{d}{x}\right.}+{c}$$
$$\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}=\int{\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}.{\left({t}-{1}\right)}^{{{4}}}{\left.{d}{t}\right.}+{c}$$
$$\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}=\int{\left({t}+{1}\right)}.{\left({t}-{1}\right)}{\left.{d}{t}\right.}+{c}$$
$$\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}=\int{\left({t}^{{{2}}}-{1}\right)}{\left.{d}{t}\right.}+{c}$$
$$\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}={\frac{{{t}^{{{3}}}}}{{{3}}}}-{t}+{c}$$
karton

Solution:
$$(t-1)^{3} \frac{ds}{dt}+4(t-1)^{2}s=t+1; t>1$$
$$\frac{ds}{dt}+\frac{4(t-1)^{2}}{(t-1)^{3}} s=\frac{t+1}{(t-1)^{3}}$$
$$\frac{ds}{dt}+\frac{4}{(t-1)}\ s=\frac{t+1}{(t-1)^{3}}$$
Compare the differential equation with $$\frac{dy}{dx}+Py=Q$$
$$P=\frac{4}{(t-1)^{2}}\ \text{and} Q=\frac{t+1}{(t-1)^{3}}$$
Solution of given differential equation is $$y. (I.F)=\int Q. (I.F)dx+c$$
$$s.(t-1)^{4}=\int \frac{t+1}{(t-1)^{3}}.(t-1)^{4}dt+c$$
$$s.(t-1)^{4}=\frac{t^{3}}{3}-t+c$$